Question

Five fair coins are tossed independently. What is the probability that at least two heads appear?

• A. $\frac{13}{16}$
• B. $\frac{7}{16}$
• C. $\frac{5}{16}$
• D. $\frac{11}{16}$

Hint:

For "at least" probability questions, always consider the complement.
Calculating $1 - [P(0) + P(1)]$ requires only two quick binomial computations, whereas direct calculation of $P(2) + P(3) + P(4) + P(5)$ requires four, saving valuable time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For independent tosses of a fair coin, we use Binomial Probability: $P(X=r) = {}^nC_r (p)^r (q)^{n-r}$, where $n=5$ and $p=q=1/2$.
Step 2: Using the Complement Rule:
$P(\text{at least 2 heads}) = 1 - [P(0 \text{ heads}) + P(1 \text{ head})]$. Total outcomes $= 2^5 = 32$.
Step 3: Calculating Probabilities:

$P(0 \text{ heads}) = {}^5C_0 (1/2)^5 = 1/32$.
$P(1 \text{ head}) = {}^5C_1 (1/2)^5 = 5/32$.
Sum $= 6/32 = 3/16$.
Step 4: Final Answer:
Probability $= 1 - 3/16 = 13/16$.